/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 The owner of a professional foot... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The owner of a professional football team believes that the league has become more offense oriented since five years ago. To check his belief, 32 randomly selected games from one year's schedule were compared to 32 randomly selected games from the schedule five years later. Since more offense produces more offensive yards per game, the owner analyzed the following information on offensive yards per game (oypg).

Short Answer

Expert verified
Conduct a t-test to compare means; if p < 0.05, offensive yards increased.

Step by step solution

01

Define the Problem

The owner wants to determine if there is a significant increase in offensive yards per game (oypg) between two different years. We will compare the means of offensive yards from both years using a statistical hypothesis test.
02

Set Up the Null and Alternative Hypotheses

Let \( \mu_1 \) be the mean offensive yards per game in the initial year, and \( \mu_2 \) be the mean offensive yards per game from five years later. The null hypothesis \( H_0 \) is \( \mu_1 = \mu_2 \), meaning there is no increase in offensive yards. The alternative hypothesis \( H_a \) is \( \mu_1 < \mu_2 \), which suggests offense has increased.
03

Collect and Summarize the Data

It's indicated that 32 games were sampled from each year, but the exact offensive yard data for these games was not provided. For this step, calculate the mean and standard deviation for each sample.
04

Choose and Conduct the Statistical Test

Assuming the data follows a normal distribution, perform a two-sample t-test to compare the means. This test helps determine if there is a significant difference between the offensive yards per game from the two years.
05

Interpret the Results

Based on the t-test result (usually expressed as a p-value), conclude whether to reject or fail to reject the null hypothesis. If the p-value is less than the significance level (commonly 0.05), reject the null hypothesis in favor of the alternative.
06

Make a Conclusion

If the null hypothesis is rejected, conclude that there is evidence supporting the increase in offensive yards over the five-year period. If not, suggest that there is not enough evidence to support the owner's belief.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Offensive Yards Per Game
Offensive yards per game (oypg) is a key metric in football used to assess a team's offensive performance in generating yardage during matches. It's calculated by dividing the total number of offensive yards gained by a team during a game by the number of games played. This statistic helps highlight how effective a team is at advancing the ball on the field. Tracking offensive yards per game gives insights into both the individual performance of players and the overall effectiveness of offensive strategies. An average fan might notice high oypg often aligns with successful teams, as it is indicative of a productive offense. The analysis of offensive yards per game over different seasons or years is crucial for understanding trends and changes in strategies, possibly reflecting shifts towards more offense-oriented approaches.
Null and Alternative Hypotheses
In hypothesis testing, the null and alternative hypotheses are fundamental concepts used to test assumptions made about a population. The null hypothesis represents the status quo or a baseline hypothesis that there is no effect or difference. It is denoted by \( H_0 \). The alternative hypothesis is what you aim to support; it asserts that there is a statistically significant effect or difference. This is symbolized as \( H_a \). In our specific study of offensive yards per game:
  • The null hypothesis \( H_0 \): The mean offensive yards per game from the two different years are equal, \( \mu_1 = \mu_2 \). This indicates there is no increase in performance over time.
  • The alternative hypothesis \( H_a \): The mean offensive yards per game has increased, \( \mu_1 < \mu_2 \). This suggests a shift towards a more robust offensive game.
These hypotheses are the foundation of statistical testing, allowing us to make informed conclusions based on collected data.
Two-Sample t-Test
A two-sample t-test is a statistical method used to determine if two sets of data are significantly different from each other. It's particularly useful when you're comparing the means of two independent samples, like in our case, the offensive yards per game from two different years. This test assumes that both samples are normally distributed and that they have similar variances. When these conditions are met, the two-sample t-test can robustly assess whether any observed differences in means are statistically significant, or simply due to random chance. The process involves computing a t-statistic, which stems from the given sample means, sample sizes, and standard deviations. This t-statistic is then used to derive a p-value, which helps decide whether to support or reject the null hypothesis. If the p-value is below a predetermined significance level (commonly set at 0.05), you reject the null hypothesis, implying a significant difference in means. Understanding how to conduct and interpret a two-sample t-test is essential for rigorous data analysis, enabling one to draw reliable conclusions from comparative studies.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years. a. Construct the \(95 \%\) confidence interval for the difference in the means based on these data. b. Test the hypothesis that there is a difference in the means against the null hypothesis that there is no difference. Use the \(1 \%\) level of significance. c. Compute the observed significance of the test in part (b).

The coaching staff of professional football team believes that the rushing offense has become increasingly potent in recent years. To investigate this belief, 20 randomly selected games from one year's schedule were compared to 11 randomly selected games from the schedule five years later. The sample information on passing yards per game (pypg) is summarized below. \begin{tabular}{|c|c|c|c|} \hline & \(n\) & \(=\) & \(s\) \\ \hline pypg previously & 20 & 203 & 38 \\ \hline pypg recently & 11 & 232 & 33 \\ \hline \end{tabular} a. Construct the \(95 \%\) confidence interval for the difference in the population means based on these data. b. Test, at the \(5 \%\) level of significance, whether the data on passing yards per game provide sufficient evidence to conclude that the passing offense has become more potent in recent years.

An educational researcher wishes to estimate the difference in average scores of elementary school children on two versions of a 100-point standardized test, at \(99 \%\) confidence and to within two points. Estimate the minimum equal sample sizes necessary if it is known that the standard deviation of scores on different versions of such tests is 4.9

A journalist plans to interview an equal number of members of two political parties to compare the proportions in each party who favor a proposal to allow citizens with a proper license to carry a concealed handgun in public parks. Let \(p_{1}\) and \(p_{2}\) be the true proportions of members of the two parties who are in favor of the proposal. Suppose it is desired to find a \(95 \%\) confidence interval for estimating \(p_{1-p 2}\) to within 0.05. Estimate the minimum equal number of members of each party that must be sampled to meet these criteria.

Estimate the common sample size \(n\) of equally sized independent samples needed to estimate \(\mu 1-\mu 2\) as specified when the population standard deviations are as shown. a. \(\quad 80 \%\) confidence, to within 2 units, \(\sigma_{1}=14\) and \(\sigma_{2}=23\) b. \(90 \%\) confidence, to within 0.3 units, \(\sigma_{1}=1.3\) and \(\sigma_{2}=0.8\) c. \(99 \%\) confidence, to within 11 units, \(\sigma_{1}=42\) and \(\sigma_{2}=37\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.